lesson 12: the volume formula of a sphere
TRANSCRIPT
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 12
GEOMETRY
Lesson 12: The Volume Formula of a Sphere
172
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Lesson 12: The Volume Formula of a Sphere
Student Outcomes
Students give an informal argument using Cavalieriโs principle for the volume formula of a sphere and use the
volume formula to derive a formula for the surface area of a sphere.
Lesson Notes
Students informally derive the volume formula of a sphere in Lesson 12 (G-GMD.A.2). To do so, they examine the
relationship between a hemisphere, cone, and cylinder, each with the same radius, and for the cone and cylinder, a
height equal to the radius. Students discover that when the solids are aligned, the sum of the area of the cross-sections
at a given height of the hemisphere and cone are equal to the area of the cross-section of the cylinder. They use this
discovery and their understanding of Cavalieriโs principle to establish a relationship between the volumes of each,
ultimately leading to the volume formula of a sphere.
Classwork
Opening Exercise (5 minutes)
Opening Exercise
Picture a marble and a beach ball. Which one would you describe as a sphere? What differences between the two could
possibly impact how we describe what a sphere is?
Answers vary; some students may feel that both are spheres. Some may distinguish that one is a solid while the other is
hollow. Share out responses before moving to the definition of sphere.
Definition Characteristics
Sphere: Given a point ๐ช in the three-dimensional space and
a number ๐ > ๐, the sphere with center ๐ช and radius ๐ is
the set of all points in space that are distance ๐ from the
point ๐ช.
Solid Sphere or Ball: Given a point ๐ช in the three-
dimensional space and a number ๐ > ๐, the solid sphere
(or ball) with center ๐ช and radius ๐ is the set of all points in
space whose distance from the point ๐ช is less than or equal
to ๐.
Sphere: hollow
Solid Sphere: filled or solid
Examples Non-Examples
Sphere: beach ball, soap bubble
Solid Sphere: marble, planet
Sphere: an egg-shaped figure
Solid Sphere: a circle
Sphere
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 12
GEOMETRY
Lesson 12: The Volume Formula of a Sphere
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It should be noted that a sphere is just the three-dimensional analog of a circle. The teacher may want to have students
compare the definitions of the two terms.
Tell students that the term hemisphere refers to a half-sphere, and solid hemisphere refers to a solid half-sphere.
Discussion (18 minutes)
Note that students find the volume of a solid sphere by first finding half the volume, the volume of a solid hemisphere,
and then doubling. Note that we often speak of the volume of a sphere, even though we really mean the volume of the
solid sphere, just as we speak of the area of a circle when we really mean the area of a disk.
Today, we will show that the sum of the volume of a solid hemisphere of radius ๐ and the volume of a right
circular cone of radius ๐ and height ๐ is the same as the volume of a right circular cylinder of radius ๐ and
height ๐ . How could we use Cavalieri's principle to do this?
Allow students a moment to share thoughts. Some students may formulate an idea about the
relationship between the marked cross-sections based on the diagram below.
Consider the following solids shown: a solid hemisphere, ๐ป; a right circular cone, ๐; and a right circular
cylinder, ๐, each with radius ๐ and height ๐ (regarding the cone and cylinder).
The solids are aligned above a base plane that contains the bases of the hemisphere and cylinder and the
vertex of the cone; the altitude of the cone is perpendicular to this plane.
A cross-sectional plane that is distance โ from the base plane intersects the three solids. What is the shape of
each cross-section? Sketch the cross-sections, and make a conjecture about their relative sizes (e.g., order
smallest to largest, and explain why).
Each cross-section is in the shape of a disk. It looks like the cross-section of the cone will be the
smallest, the cross-section of the cylinder will be the largest, and the cross-section of the hemisphere
will be between the sizes of the other two.
Let ๐ท1, ๐ท2, ๐ท3 be the cross-sectional disks for the solid hemisphere, the cone, and the cylinder, respectively.
Let ๐1, ๐2, ๐3 be the radii of ๐ท1, ๐ท2, ๐ท3, respectively.
Our first task in order to accomplish our objective is to find the area of each cross-sectional disk. Since the
radii are all of different lengths, we want to try and find the area of each disk in terms of ๐ and โ, which are
common between the solids.
Hemisphere ๐ฏ Cone ๐ป Cylinder ๐บ
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 12
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Lesson 12: The Volume Formula of a Sphere
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Examine the hemisphere more closely in Figure 1.
What is the area formula for disk ๐ท1 in terms of ๐1?
Area(๐ท1) = ๐๐12
How can we find ๐1 in terms of โ and ๐ ?
Allow students time to piece together that the diagram they need to focus on looks like
the following figure. Take student responses before confirming with the solution.
By the Pythagorean theorem:
๐12 + โ2 = ๐ 2
๐1 = โ๐ 2 โ โ2
Once we have ๐1, substitute it into the area formula for disk ๐ท1.
The area of ๐ท1:
Area(๐ท1) = ๐๐12
Area(๐ท1) = ๐ (โ๐ 2 โ โ2)2
Area(๐ท1) = ๐๐ 2 โ ๐โ2
Let us pause and summarize what we know so far. Describe what we have shown so far.
We have shown that the area of the cross-sectional disk of the hemisphere is ๐๐ 2 โ ๐โ2.
Record this result in the classroom.
Continuing on with our goal of finding the area of each disk, now find the radius ๐2 and the area of ๐ท2 in terms
of ๐ and โ. Examine the cone more closely in Figure 2.
Figure 1
Scaffolding:
Students may have
difficulty seeing that the
hypotenuse of the right
triangle with legs โ and ๐1
is ๐ . Remind students that
โ and ๐1 meet at 90ยฐ.
Students may try to form a
trapezoid out of the
existing segments:
Nudge students to
reexamine Figure 1 and
observe that the distance
from the center of the
base disk to where ๐1
meets the hemisphere is
simply the radius of the
hemisphere.
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GEOMETRY
Lesson 12: The Volume Formula of a Sphere
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If students require a prompt, remind them that both the radius and the height of the cone are each length ๐ .
By using similar triangles:
๐2
โ=
๐
๐ = 1, or ๐2 = โ
The area of ๐ท2:
Area(๐ท2) = ๐๐22 = ๐โ2
Let us pause again and summarize what we know about the area of the cross-section of the cone. Describe
what we have shown.
We have shown that the area of the cross-sectional disk of the hemisphere is ๐๐ 2 โ ๐โ2.
Record the response next to the last summary.
Lastly, we need to find the area of disk ๐ท3 in terms of ๐ and โ. Examine the cone more closely in Figure 3.
In the case of this cylinder, will โ play a part in the area formula of disk ๐ท3? Why?
The radius ๐3 is equal to ๐ , so the area formula will not require โ this time.
The area of ๐ท3:
Area(๐ท3) = ๐๐ 2
Figure 2
Figure 3
Scaffolding:
Consider showing students the
following image as a prompt
before continuing with the
solution:
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 12
GEOMETRY
Lesson 12: The Volume Formula of a Sphere
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Write all three areas on the board while asking the following questions:
What do you now know about the three areas of the cross-sections?
Area(๐ท1) = ๐๐ 2 โ ๐โ2
Area(๐ท2) = ๐โ2
Area(๐ท3) = ๐๐ 2
Do you notice a relationship between the areas of ๐ท1, ๐ท2, ๐ท3? What is it?
The area of ๐ท3 is the sum of the areas of ๐ท1 and ๐ท2.
๐ด๐๐๐(๐ท1) + ๐ด๐๐๐(๐ท2) = ๐ด๐๐๐(๐ท3)
(๐๐ 2 โ ๐โ2) + ๐โ2 = ๐๐ 2
Then, let us review two key facts: (1) The three solids all have the same height; (2) at any given height, the
sum of the areas of the cross-sections of the hemisphere and cone are equal to the cross-section of the
cylinder.
Allow students to wrestle and share out ideas:
How does this relate to our original objective of showing that the sum of the volume of a solid hemisphere ๐ป
and the volume of cone ๐ is equal to the volume of cylinder ๐?
What does Cavalieriโs principle tell us about solids with equal heights and with cross-sections with equal areas
at any given height?
Solids that fit that criteria must have equal volumes.
By Cavalieriโs principle, we can conclude that the sum of the volumes of the hemisphere and cone are equal to
the volume of the cylinder.
Every plane parallel to the base plane intersects ๐ป โช ๐ and ๐ in cross-sections of equal area. Cavalieriโs
principle states that if every plane parallel to the two planes intersects both solids in cross-sections of equal
area, then the volumes of the two solids are equal. Therefore, the volumes of ๐ป โช ๐ and ๐ are equal.
Example (8 minutes)
Students now calculate the volume of a sphere with radius ๐ using the relationship they discovered between a
hemisphere, cone, and cylinder of radius ๐ and height ๐ .
Example
Use your knowledge about the volumes of cones and cylinders to find a volume for a solid hemisphere of radius ๐น.
We have determined that the volume of ๐ป โช ๐ is equal to the volume of ๐. What is the volume of ๐?
Vol(๐) =13
ร area of base ร height
Vol(๐) =13
๐๐ 3
What is the volume of ๐?
Vol(๐) = area of base ร height
Vol(๐) = ๐๐ 3
MP.7
MP.1
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 12
GEOMETRY
Lesson 12: The Volume Formula of a Sphere
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Set up an equation and solve for Vol(๐ป).
Vol(๐ป) +13
๐๐ 3 = ๐๐ 3
Vol(๐ป) =2
3๐๐ 3
What does 2
3๐๐ 3 represent?
The Vol(๐ป), or the volume of a hemisphere with radius ๐
Then, what is the volume formula for a whole sphere?
Twice the volume of a solid hemisphere or 4
3๐๐ 3
The volume formula for a sphere is ๐ =43
๐๐ 3.
Exercises (8 minutes)
Have students complete any two of the exercises.
Exercises
1. Find the volume of a sphere with a diameter of ๐๐ ๐๐ฆ to one decimal place.
๐ฝ =๐
๐๐ (๐)๐
๐ฝ =๐
๐๐ (๐๐๐)
๐ฝ โ ๐๐๐. ๐
The volume of the sphere is approximately ๐๐๐. ๐ ๐๐ฆ๐.
2. An ice cream cone is ๐๐ ๐๐ฆ deep and ๐ ๐๐ฆ across the opening of the cone. Two hemisphere-shaped scoops of ice
cream, which also have diameters of ๐ ๐๐ฆ, are placed on top of the cone. If the ice cream were to melt into the
cone, would it overflow?
๐๐จ๐ฅ๐ฎ๐ฆ๐(๐๐จ๐ง๐) =๐
๐๐ (
๐
๐)
๐
โ (๐๐) ๐๐จ๐ฅ๐ฎ๐ฆ๐(๐ข๐๐ ๐๐ซ๐๐๐ฆ) =๐
๐๐ (
๐
๐)
๐
๐๐จ๐ฅ๐ฎ๐ฆ๐(๐๐จ๐ง๐) โ ๐๐ ๐๐จ๐ฅ๐ฎ๐ฆ๐(๐ข๐๐ ๐๐ซ๐๐๐ฆ) โ ๐๐. ๐
The volume the cone can hold is roughly ๐๐ ๐๐ฆ๐. The volume of ice cream is roughly ๐๐. ๐ ๐๐ฆ๐.
The melted ice cream will not overflow because there is significantly less ice cream than there is space in the cone.
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 12
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Lesson 12: The Volume Formula of a Sphere
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3. Bouncy rubber balls are composed of a hollow rubber shell ๐. ๐" thick and an outside diameter of ๐. ๐". The price of
the rubber needed to produce this toy is $๐. ๐๐๐/๐ข๐ง๐.
a. What is the cost of producing ๐ case, which holds ๐๐ such balls? Round to the nearest cent.
The outer shell of the ball is ๐. ๐" thick, so the hollow center has a diameter of ๐. ๐" and, therefore, a radius of
๐. ๐".
The volume of rubber in a ball is equal to the difference of the volumes of the entire sphere and the hollow
center.
๐ฝ = [๐
๐๐ (๐. ๐)๐] โ [
๐
๐๐ (๐. ๐)๐]
๐ฝ =๐
๐๐ [(๐. ๐)๐ โ (๐. ๐)๐]
๐ฝ =๐
๐๐ [๐. ๐๐๐]
๐ฝ โ ๐. ๐๐๐๐
The volume of rubber needed for each individual ball is approximately ๐. ๐๐๐๐ ๐ข๐ง๐. For a case of ๐๐ rubber
balls, the volume of rubber required is ๐
๐๐ [๐. ๐๐๐] โ ๐๐ โ ๐๐. ๐, or ๐๐. ๐ ๐ข๐ง๐.
Total cost equals the cost per cubic inch times the total volume for the case.
๐๐จ๐ญ๐๐ฅ ๐๐จ๐ฌ๐ญ = [๐. ๐๐๐] โ [๐
๐๐ [๐. ๐๐๐] โ ๐๐]
๐๐จ๐ญ๐๐ฅ ๐๐จ๐ฌ๐ญ โ ๐. ๐๐
The total cost for rubber to produce a case of ๐๐ rubber balls is approximately $๐. ๐๐.
b. If each ball is sold for $๐. ๐๐, how much profit is earned on each ball sold?
๐๐จ๐ญ๐๐ฅ ๐ฌ๐๐ฅ๐๐ฌ = (๐ฌ๐๐ฅ๐ฅ๐ข๐ง๐ ๐ฉ๐ซ๐ข๐๐) โ (๐ฎ๐ง๐ข๐ญ๐ฌ ๐ฌ๐จ๐ฅ๐)
๐๐จ๐ญ๐๐ฅ ๐ฌ๐๐ฅ๐๐ฌ = (๐. ๐)(๐๐)
๐๐จ๐ญ๐๐ฅ ๐ฌ๐๐ฅ๐๐ฌ = ๐
The total sales earned for selling the case of rubber balls is $๐.
๐๐ซ๐จ๐๐ข๐ญ ๐๐๐ซ๐ง๐๐ = (๐ญ๐จ๐ญ๐๐ฅ ๐ฌ๐๐ฅ๐๐ฌ) โ (๐๐ฑ๐ฉ๐๐ง๐ฌ๐)
๐๐ซ๐จ๐๐ข๐ญ ๐๐๐ซ๐ง๐๐ = (๐) โ (๐. ๐๐)
๐๐ซ๐จ๐๐ข๐ญ ๐๐๐ซ๐ง๐๐ = ๐. ๐๐
The total profit earned after selling the case of rubber balls is $๐. ๐๐.
๐๐ซ๐จ๐๐ข๐ญ ๐ฉ๐๐ซ ๐ฎ๐ง๐ข๐ญ =๐. ๐๐
๐๐
๐๐ซ๐จ๐๐ข๐ญ ๐ฉ๐๐ซ ๐ฎ๐ง๐ข๐ญ = ๐. ๐๐๐๐
The profit earned on the sale of each ball is approximately $๐. ๐๐.
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 12
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Lesson 12: The Volume Formula of a Sphere
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Extension: In the Extension, students derive the formula for the surface area of a sphere from the volume formula of
the sphere.
Extension
The volume formula for a solid sphere can be used to find the surface area of a sphere with radius ๐ .
Cover the sphere with non-overlapping regions with area ๐ด1, ๐ด2, ๐ด3, โฆ , ๐ด๐.
For each region, draw a โconeโ with the region for a base and the center of the sphere for the vertex (see
image).
One says โconeโ since the base is not flat but rather a piece of a sphere. However, it is an approximation of a cone.
The volume of the cone with base area ๐ด1 and height equal to the radius ๐ is approximately
Volume(๐ด1) โ1
3โ ๐ด1 โ ๐ .
How is the volume of the sphere related to the volume of the cones? Explain and then show what you mean in
a formula.
The volume of the sphere is the sum of the volumes of these โcones.โ
4
3๐๐ 3 โ
1
3โ (๐ด1 + ๐ด2 + โฏ + ๐ด๐) โ ๐
Let ๐ be the surface area of the sphere. Then, the sum is ๐ด1 + ๐ด2 + โฏ + ๐ด๐ = ๐. We can rewrite the last
approximation as
4
3๐๐ 3 โ
1
3โ ๐ โ ๐ .
Remind students that the right-hand side is only approximately equal to the left-hand side because the bases of these
โconesโ are slightly curved, and ๐ is not the exact height.
As the regions are made smaller, and we take more of them, the โconesโ more closely approximate actual
cones. Hence, the volume of the solid sphere would actually approach 1
3โ ๐ โ ๐ as the number of regions
approaches โ.
Under this assumption, we will use the equal sign instead of the approximation symbol:
4
3๐๐ 3 =
1
3โ ๐ โ ๐ .
MP.7
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Lesson 12: The Volume Formula of a Sphere
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Solving for ๐, we get
๐ = 4๐๐ 2.
Thus, the formula for the surface area of a sphere is
Surface Area = 4๐๐ 2.
Closing (1 minute)
What is the relationship between a hemisphere, a cone, and a cylinder, all of which have the same radius, if
the height of the cone and cylinder is equal to the radius?
The area of the cross-sections taken at height โ of the hemisphere and cone is equal to the area of the
cross-section of the cylinder taken at the same height. By Cavalieriโs principle, we can conclude that the
sum of the volumes of the hemisphere and cone is equal to the volume of the cylinder.
What is the volume formula of a sphere?
๐ =43
๐๐ 3
Exit Ticket (5 minutes)
Lesson Summary
SPHERE: Given a point ๐ช in the three-dimensional space and a number ๐ > ๐, the sphere with center ๐ช and radius ๐
is the set of all points in space that are distance ๐ from the point ๐ช.
SOLID SPHERE OR BALL: Given a point ๐ช in the three-dimensional space and a number ๐ > ๐, the solid sphere (or ball)
with center ๐ช and radius ๐ is the set of all points in space whose distance from the point ๐ช is less than or equal to ๐.
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Lesson 12: The Volume Formula of a Sphere
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Name Date
Lesson 12: The Volume Formula of a Sphere
Exit Ticket
1. Snow globes consist of a glass sphere that is filled with liquid and other contents. If
the inside radius of the snow globe is 3 in., find the approximate volume of material
in cubic inches that can fit inside.
2. The diagram shows a hemisphere, a circular cone, and a circular cylinder with heights and radii both equal to 9.
a. Sketch parallel cross-sections of each solid at height 3 above plane ๐.
b. The base of the hemisphere, the vertex of the cone, and the base of the cylinder lie in base plane ๐ฒ. Sketch
parallel cross-sectional disks of the figures at a distance โ from the base plane, and then describe how the
areas of the cross-sections are related.
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Lesson 12: The Volume Formula of a Sphere
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Exit Ticket Sample Solutions
1. Snow globes consist of a glass sphere that is filled with liquid and other contents. If
the inside radius of the snow globe is ๐ ๐ข๐ง., find the approximate volume of material in
cubic inches that can fit inside.
๐๐จ๐ฅ๐ฎ๐ฆ๐ =๐
๐๐ ๐น๐
๐๐จ๐ฅ๐ฎ๐ฆ๐ =๐
๐๐ โ ๐๐
๐๐จ๐ฅ๐ฎ๐ฆ๐ = ๐๐๐ โ ๐๐๐. ๐
The snow globe can contain approximately ๐๐๐. ๐ ๐ข๐ง๐ of material.
2. The diagram shows a hemisphere, a circular cone, and a circular cylinder with heights and radii both equal to ๐.
a. Sketch parallel cross-sections of each solid at height ๐ above plane ๐ท.
See the diagram above.
b. The base of the hemisphere, the vertex of the cone, and the base of the cylinder lie in base plane ๐ฌ. Sketch
parallel cross-sectional disks of the figures at a distance ๐ from the base plane, and then describe how the
areas of the cross-sections are related.
The cross-sectional disk is ๐
๐ of the distance from plane ๐ท to the base of the cone, so its radius is also
๐
๐ of the
radius of the coneโs base. Therefore, the radius of the disk is ๐, and the disk has area ๐๐ .
The area of the cross-sectional disk in the cylinder is the same as the area of the cylinderโs base since the disk
is congruent to the base. The area of the disk in the cylinder is ๐๐๐ .
๐๐ซ๐๐(๐๐ฒ๐ฅ๐ข๐ง๐๐๐ซ ๐๐ข๐ฌ๐ค) = ๐๐ซ๐๐(๐ก๐๐ฆ๐ข๐ฌ๐ฉ๐ก๐๐ซ๐ ๐๐ข๐ฌ๐ค) + ๐๐ซ๐๐(๐๐จ๐ง๐ ๐๐ข๐ฌ๐ค)
๐๐๐ = ๐๐ซ๐๐(๐ก๐๐ฆ๐ข๐ฌ๐ฉ๐ก๐๐ซ๐ ๐๐ข๐ฌ๐ค) + ๐๐
๐๐๐ = ๐๐ซ๐๐(๐ก๐๐ฆ๐ข๐ฌ๐ฉ๐ก๐๐ซ๐ ๐๐ข๐ฌ๐ค)
The area of the disk in the hemisphere at height ๐ above plane ๐ท is ๐๐๐ or approximately ๐๐๐. ๐.
Problem Set Sample Solutions
1. A solid sphere has volume ๐๐๐ . Find the radius of the sphere.
๐
๐๐ ๐๐ = ๐๐๐
๐๐ = ๐๐
๐ = ๐
Therefore, the radius is ๐ units.
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 12
GEOMETRY
Lesson 12: The Volume Formula of a Sphere
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2. A sphere has surface area ๐๐๐ . Find the radius of the sphere.
๐๐ ๐๐ = ๐๐๐
๐๐ = ๐
๐ = ๐
Therefore, the radius is ๐ units.
3. Consider a right circular cylinder with radius ๐ and height ๐. The area of each base is ๐ ๐๐. Think of the lateral
surface area as a label on a soup can. If you make a vertical cut along the label and unroll it, the label unrolls to the
shape of a rectangle.
a. Find the dimensions of the rectangle.
๐๐ ๐ by ๐
b. What is the lateral (or curved) area of the cylinder?
๐๐๐ญ๐๐ซ๐๐ฅ ๐๐ซ๐๐ = ๐ฅ๐๐ง๐ ๐ญ๐ก ร ๐ฐ๐ข๐๐ญ๐ก
๐๐๐ญ๐๐ซ๐๐ฅ ๐๐ซ๐๐ = ๐๐ ๐๐
4. Consider a right circular cone with radius ๐, height ๐, and slant height ๐ (see Figure 1). The area of the base is ๐ ๐๐.
Open the lateral area of the cone to form part of a disk (see Figure 2). The surface area is a fraction of the area of
this disk.
a. What is the area of the entire disk in Figure 2?
๐๐ซ๐๐ = ๐ ๐๐
b. What is the circumference of the disk in Figure 2?
๐๐ข๐ซ๐๐ฎ๐ฆ๐๐๐ซ๐๐ง๐๐ = ๐๐ ๐
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 12
GEOMETRY
Lesson 12: The Volume Formula of a Sphere
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The length of the arc on this circumference (i.e., the arc that borders the green region) is the circumference of the
base of the cone with radius ๐ or ๐๐ ๐. (Remember, the green region forms the curved portion of the cone and
closes around the circle of the base.)
c. What is the ratio of the area of the disk that is shaded to the area of the whole disk?
๐๐ ๐
๐๐ ๐=
๐
๐
d. What is the lateral (or curved) area of the cone?
๐
๐ร ๐ ๐๐ = ๐ ๐๐
5. A right circular cone has radius ๐ ๐๐ฆ and height ๐ ๐๐ฆ. Find the lateral surface area.
By the Pythagorean theorem, the slant height of the cone is ๐ ๐๐ฆ.
The lateral surface area in ๐๐ฆ๐ is ๐ ๐๐ = ๐ ๐ โ ๐ = ๐๐๐ . The lateral surface area is ๐๐๐ ๐๐ฆ๐.
6. A semicircular disk of radius ๐ ๐๐ญ. is revolved about its diameter (straight side) one complete revolution. Describe
the solid determined by this revolution, and then find the volume of the solid.
The solid is a solid sphere with radius ๐ ๐๐ญ. The volume in ๐๐ญ๐ is ๐
๐๐ ๐๐ = ๐๐๐ . The volume is ๐๐๐ ๐๐ญ๐.
7. A sphere and a circular cylinder have the same radius, ๐, and the height of the cylinder is ๐๐.
a. What is the ratio of the volumes of the solids?
The volume of the sphere is ๐
๐๐ ๐๐, and the volume of the cylinder is ๐๐ ๐๐, so the ratio of the volumes is the
following:
๐๐
๐ ๐๐
๐๐ ๐๐=
๐
๐=
๐
๐.
The ratio of the volume of the sphere to the volume of the cylinder is ๐: ๐.
b. What is the ratio of the surface areas of the solids?
The surface area of the sphere is ๐๐ ๐๐, and the surface area of the cylinder is ๐(๐ ๐๐) + (๐๐ ๐ โ ๐๐) = ๐๐ ๐๐,
so the ratio of the surface areas is the following:
๐๐ ๐๐
๐๐ ๐๐=
๐
๐=
๐
๐.
The ratio of the surface area of the sphere to the surface area of the cylinder is ๐: ๐.
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 12
GEOMETRY
Lesson 12: The Volume Formula of a Sphere
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8. The base of a circular cone has a diameter of ๐๐ ๐๐ฆ and an altitude of ๐๐ ๐๐ฆ. The cone is filled with water. A
sphere is lowered into the cone until it just fits. Exactly one-half of the sphere remains out of the water. Once the
sphere is removed, how much water remains in the cone?
๐ฝ๐๐จ๐ง๐ =๐
๐๐ (๐)๐(๐๐)
๐ฝ๐ก๐๐ฆ๐ข๐ฌ๐ฉ๐ก๐๐ซ๐ = (๐
๐)
๐
๐๐ (๐)๐
๐ฝ๐๐จ๐ง๐ โ ๐ฝ๐ก๐๐ฆ๐ข๐ฌ๐ฉ๐ก๐๐ซ๐ =๐๐
๐๐ (๐)๐ โ
๐
๐๐ (๐)๐
๐ฝ๐๐จ๐ง๐ โ ๐ฝ๐ก๐๐ฆ๐ข๐ฌ๐ฉ๐ก๐๐ซ๐ =๐
๐๐ (๐)๐[๐๐ โ ๐]
๐ฝ๐๐จ๐ง๐ โ ๐ฝ๐ก๐๐ฆ๐ข๐ฌ๐ฉ๐ก๐๐ซ๐ =๐
๐๐ (๐)๐
The volume of water that remains in the cone is ๐
๐๐ (๐)๐, or approximately ๐, ๐๐๐. ๐ ๐๐ฆ๐.
9. Teri has an aquarium that is a cube with edge lengths of ๐๐ inches. The aquarium is ๐
๐ full of water. She has a
supply of ball bearings, each having a diameter of ๐
๐ inch.
a. What is the maximum number of ball bearings that Teri can drop into the aquarium without the water
overflowing?
๐๐ ๐ข๐ง. ร ๐๐ ๐ข๐ง. ร ๐๐ ๐ข๐ง. = ๐๐๐๐๐ ๐ข๐ง๐
The volume of the cube is ๐๐, ๐๐๐ ๐ข๐ง๐.
๐๐ ๐ข๐ง. ร ๐๐ ๐ข๐ง. ร ๐๐ ๐ข๐ง. = ๐๐๐๐๐ ๐ข๐ง๐
The volume of water in the cube is ๐๐, ๐๐๐ ๐ข๐ง๐.
๐๐๐๐๐ ๐ข๐ง๐ โ ๐๐๐๐๐ ๐ข๐ง๐ = ๐๐๐๐ ๐ข๐ง๐
The remaining volume in the aquarium is ๐, ๐๐๐ ๐ข๐ง๐.
๐ฝ =๐
๐๐ (
๐
๐)
๐
๐ฝ =๐
๐๐ (
๐๐
๐๐๐)
๐ฝ =๐
๐๐๐๐
The volume of one ball bearing is ๐
๐๐๐๐ ๐ข๐ง๐ or approximately ๐. ๐๐๐ ๐ข๐ง๐.
๐๐๐๐
๐๐๐๐
๐ =
๐๐๐๐๐
๐ โ ๐๐๐๐๐. ๐
Teri can drop ๐๐, ๐๐๐ ball bearings into the aquarium without the water overflowing. If she drops in one
more, the water (theoretically without considering a meniscus) will overflow.
b. Would your answer be the same if the aquarium were ๐๐
full of sand? Explain.
In the original problem, the water will fill the gaps between the ball bearings as they are dropped in;
however, the sand will not fill the gaps unless the mixture of sand and ball bearings is continuously stirred.
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 12
GEOMETRY
Lesson 12: The Volume Formula of a Sphere
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c. If the aquarium were empty, how many ball bearings would fit on the
bottom of the aquarium if you arranged them in rows and columns, as
shown in the picture?
The length and width of the aquarium are ๐๐ inches, and ๐๐ inches divided
into ๐
๐ inch intervals is ๐๐, so each row and column would contain
๐๐ bearings. The total number of bearings in a single layer would be ๐, ๐๐๐.
d. How many of these layers could be stacked inside the aquarium without going over the top of the aquarium?
How many bearings would there be altogether?
The aquarium is ๐๐ inches high as well, so there could be ๐๐ layers of bearings for a total of ๐๐, ๐๐๐
bearings.
e. With the bearings still in the aquarium, how much water can be poured into the aquarium without
overflowing?
(๐๐๐๐๐) (๐
๐๐๐๐ ) โ ๐๐๐๐. ๐๐๐โ๐๐๐
The total volume of the ball bearings is approximately ๐๐๐๐. ๐๐๐โ๐๐๐ ๐ข๐ง๐.
๐๐๐๐๐ โ ๐๐๐๐. ๐๐๐โ๐๐๐ = ๐๐๐๐. ๐๐๐โ๐๐๐
The space between the ball bearings has a volume of ๐๐๐๐. ๐๐๐โ๐๐๐ ๐ข๐ง๐.
๐๐๐๐. ๐๐๐โ๐๐๐ ร ๐. ๐๐๐โ๐๐๐ = ๐๐. ๐๐๐โ๐๐๐โ๐๐
With the bearings in the aquarium, approximately ๐๐. ๐ gallons of water could be poured in without
overflowing the tank.
f. Approximately how much of the aquarium do the ball bearings occupy?
A little more than half of the space.
10. Challenge: A hemispherical bowl has a radius of ๐ meters. The bowl is filled with water to a depth of ๐ meter.
What is the volume of water in the bowl? (Hint: Consider a cone with the same base radius and height and the
cross-section of that cone that lies ๐ meter from the vertex.)
The volume of a hemisphere with radius ๐ is equal to the
difference of the volume of a circular cylinder with radius ๐
and height ๐ and the volume of a circular cone with base
radius ๐ and height ๐. The cross-sections of the circular cone
and the hemisphere taken at the same height ๐ from the
vertex of the cone and the circular face of the hemisphere
have a sum equal to the base of the cylinder.
Using Cavalieriโs principle, the volume of the water that
remains in the bowl can be found by calculating the volume
of the circular cylinder with the circular cone removed, and
below the cross-section at a height of ๐ (See diagram right).
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 12
GEOMETRY
Lesson 12: The Volume Formula of a Sphere
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The area of the base of the cone, hemisphere, and cylinder:
๐๐ซ๐๐ = ๐ ๐๐
๐๐ซ๐๐ = ๐ (๐ ๐ฆ)๐ = ๐๐ ๐ฆ๐
The height of the given cross-section is ๐ ๐ฆ =๐๐
(๐ ๐ฆ), so the scale factor of the
radius of the cross-sectional disk in the cone is ๐
๐, and the area is then (
๐๐
)๐
ร (๐๐ ๐ฆ๐) = ๐ ๐ฆ๐.
The total volume of the cylinder: The total volume of the cone:
๐ฝ = ๐ฉ๐ ๐ฝ =๐๐
๐ ๐๐
๐ฝ = ๐ (๐ ๐ฆ)๐ โ ๐ ๐ฆ ๐ฝ =๐๐
๐ (๐ ๐ฆ)๐
๐ฝ = ๐๐ ๐ฆ๐ ๐ฝ =๐๐
๐ ๐ฆ๐
The volume of the cone above the cross-section:
๐ฝ =๐
๐๐ (๐ ๐ฆ)๐ =
๐
๐ ๐ฆ๐
The volume of the cone below the cross-section:
๐ฝ =๐
๐๐ ๐ฆ๐ โ
๐
๐๐ ๐ฆ๐ =
๐
๐๐ ๐ฆ๐
The volume of the cylinder below the cross-section is ๐๐ ๐ฆ๐.
The volume of the cylinder below the cross-section with the section of cone removed:
๐ฝ = ๐๐ ๐ฆ๐ โ๐
๐๐ ๐ฆ๐ =
๐
๐๐ ๐ฆ๐
The volume of water left in the bowl is (๐๐
๐ ) ๐ฆ๐, or approximately ๐. ๐ ๐ฆ๐.
11. Challenge: A certain device must be created to house a scientific instrument. The housing must be a spherical shell,
with an outside diameter of ๐ ๐ฆ. It will be made of a material whose density is ๐๐ ๐ /๐๐ฆ๐. It will house a sensor
inside that weighs ๐. ๐ ๐ค๐ . The housing, with the sensor inside, must be neutrally buoyant, meaning that its density
must be the same as water. Ignoring any air inside the housing, and assuming that water has a density of ๐ ๐ /๐๐ฆ๐,
how thick should the housing be made so that the device is neutrally buoyant? Round your answer to the nearest
tenth of a centimeter.
Volume of outer sphere: ๐
๐๐ (๐๐)๐
Volume of inner sphere: ๐
๐๐ (๐)๐
Volume of shell: ๐
๐๐ [(๐๐)๐ โ (๐)๐]
Mass of shell:
๐๐๐ฌ๐ฌ = (๐๐) (๐
๐๐ [(๐๐)๐ โ (๐)๐])
In order for the device to be neutrally buoyant:
๐๐๐ง๐ฌ๐ข๐ญ๐ฒ ๐จ๐ ๐ฐ๐๐ญ๐๐ซ =๐ฆ๐๐ฌ๐ฌ(๐ฌ๐๐ง๐ฌ๐จ๐ซ) + ๐ฆ๐๐ฌ๐ฌ(๐ฌ๐ก๐๐ฅ๐ฅ)
๐ฏ๐จ๐ฅ๐ฎ๐ฆ๐ ๐จ๐ ๐ก๐จ๐ฎ๐ฌ๐ข๐ง๐
๐ =๐๐๐๐ + (๐๐) (
๐๐
๐ [(๐๐)๐ โ (๐)๐])
๐๐
๐ (๐๐)๐
๐ = โ๐๐๐๐ + (๐๐)(๐๐๐)
๐๐
๐
The thickness of the shell is ๐๐ โ โ๐๐๐๐ +(๐๐)(๐๐๐
)
๐๐
๐
, which is approximately ๐. ๐ ๐๐ฆ.
NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 12
GEOMETRY
Lesson 12: The Volume Formula of a Sphere
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12. Challenge: An inverted, conical tank has a circular base of radius ๐ ๐ฆ and a height of ๐ ๐ฆ and is full of water. Some
of the water drains into a hemispherical tank, which also has a radius of ๐ ๐ฆ. Afterward, the depth of the water in
the conical tank is ๐๐ ๐๐ฆ. Find the depth of the water in the hemispherical tank.
Total volume of water:
๐ฝ =๐
๐๐ (๐)๐(๐)
๐ฝ =๐๐
๐ ; the total volume of water is ๐
๐๐ ๐ฆ๐.
The cone formed when the water level is at ๐๐ ๐๐ฆ (๐. ๐ ๐ฆ) is similar to the original
cone; therefore, the radius of that cone is also ๐๐ ๐๐ฆ.
The volume of water once the water level drops to ๐๐ ๐๐ฆ:
๐ฝ =๐
๐๐ (๐. ๐)๐(๐. ๐ ๐ฆ)
๐ฝ =๐
๐๐ (๐. ๐๐๐)
The volume of water in the cone water once the water level drops to ๐๐ ๐๐ฆ is ๐
๐๐ (๐. ๐๐๐) ๐ฆ๐.
The volume of water that has drained into the hemispherical tank:
๐ฝ = [๐
๐๐ โ
๐
๐๐ (๐. ๐๐๐)]
๐ฝ =๐
๐๐ [๐ โ (๐. ๐๐๐)]
๐ฝ =๐
๐๐ [๐. ๐๐๐]
๐ฝ = ๐. ๐๐๐๐
The volume of water in the hemispherical tank is ๐. ๐๐๐๐ ๐ฆ๐.
The pool of water that sits in the hemispherical tank is also hemispherical.
๐. ๐๐๐๐ = (๐
๐)
๐
๐๐ ๐ซ๐
๐ซ๐ = ๐. ๐๐๐ (๐
๐)
๐ซ๐ = ๐. ๐๐๐
๐ โ ๐. ๐๐
The depth of the water in the hemispherical tank is approximately ๐. ๐๐ ๐ฆ.